Let $f(x)=\dfrac{-6x^2+x-1}{x^2+3}$. Find $\lim_{x\to\infty}f(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{1}{3}$ (Choice B) B $-6$ (Choice C) C $0$ (Choice D) D The limit is unbounded
Solution: Limits of rational functions are determined according to three cases: If the degree of the numerator is less than the degree of the denominator, then the limit is $0$. If the degree of the numerator is equal to the degree of the denominator, then the limit is the ratio of the leading coefficients. If the degree of the numerator is greater than the degree of the denominator, then the limit is unbounded. In the case of $f(x)=\dfrac{-6x^2+x-1}{x^2+3}$, the degree of the numerator $(2)$ is equal to the degree of the denominator $(2)$. Therefore, the limit is the ratio of the leading coefficients, which are $-6$ and $1$. In other words, $\lim_{x\to\infty}f(x)=\dfrac{-6}{1}=-6$. The limit can also be found directly, without memorizing the three cases, as done here. This calculation uses the fact that for any nonzero number $k$ and positive power $n$, the limit $\lim_{x\to\infty}\dfrac{k}{x^n}$ is equal to $0$. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}f(x) \\\\ &=\lim_{x\to\infty}\dfrac{-6x^2+x-1}{x^2+3} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{-6x^2+x-1}{x^{2}}}{\dfrac{x^2+3}{x^{2}}} \gray{\text{Divide both sides by }x^{2}} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{-6\cancel{x^2}}{\cancel{x^2}}+\dfrac{\cancel x}{\cancel x\cdot x}-\dfrac{\cancel 1}{ x^2}}{\dfrac{\cancel{x^2}}{\cancel{x^2}}+\dfrac{3}{x^2}} \\\\ &=\lim_{x\to\infty}\dfrac{-6+\dfrac{1}{x}-\dfrac{1}{x^2}}{1+\dfrac{3}{x^2}} \\\\ &=\dfrac{-6+0-0}{1+0} \gray{\lim_{x\to\infty}\dfrac{k}{x^n}=0} \\\\ &=-6 \end{aligned}$ In conclusion, $\lim_{x\to\infty}f(x)=-6$.